Transcript Chapter 1: Inductive Sets of Data

Chapter 1: Inductive Sets of Data


Recall definition of list

'() is a list.
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If l is a list and a is any object, then (cons a l) is a list
More generally:
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Define a simplest element
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Define remaining elements by induction
Inductive Sets of Data
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Can define (natural) numbers this way:
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0 is a number
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If i is a number, then (s i) is a number
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E.g., 3 is (s (s (s 0))) – “Church numerals”
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Russell & Whitehead (1925): try to build all of arithmethic
like this (fails because of Gödel's Incompleteness theorem)
Backus-Nauer Form (BNF)
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Simplifies description of inductive data types
<list-of-numbers> ::= 0
<list-of-numbers> ::= (<number> . <list-of-numbers>)
[where (a . b) is like (cons a b)]
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Consists of
Terminals:
0
)
. (
Non-terminals: <list-of-numbers>
Productions: <list-of-numbers> ::= ()
BNF
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Can only describe context-free structures; e.g.,
<bin-search-tree> ::=
(<key> . <bin-search-tree> <bin-search-tree>)
is inadequate for binary search trees:
7
5
3
9
6 8
11
which require a context-sensitive description
BNF: Notation
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Disjunction (this or that): |
<number> ::= <even-number> | <odd-number>
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Kleene Star (zero or more): {}*
<list-of-numbers> ::= ({<number>}*)
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Kleene Plus (one or more): {}+
<nonempty-list-of-numbers> ::= ({<number>}+
Induction
What can we do with these (BNF) inductive definitions?
1) Prove theorems about data structures. E.g. given
<bintree> ::= <symbol> | ( <bintree> <bintree> )
prove that a bintree always has an even # of parens
Induction
Basis: A tree derived from <bintree> ::= <symbol> has
zero parens. Zero is even.
Inductive step: Assume that the relation holds for two bin
trees b1, with 2m ≥ 0 parens, and b2, with 2n ≥ 0 parens.
Using the rule <bintree> ::= (<bintree> <bintree>) we make a
new bin tree b3 = (b1 b2 ) from these, which has length
2m+2n+2 = 2[m+n+1], which is even. There is no other
way to make a bin tree. Hence the relation holds for any
bin tree with zero or more parens, Q.E.D.
Induction
What can we do with these (BNF) inductive definitions?
2) Write programs that manipulate inductive data
; count parentheses in a binary tree
(define count-bt-parens
(lambda (bt)
(if (atom? bt)
0
; base case
(+ (count-bt-parens (car bt))
(count-bt-parens (cadr bt))
2)))) ; inductive step
Important Points
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Program mirrors BNF description mirrors proof.
(if (atom? bt)
0
<bintree> ::= <symbol>
Basis: A tree derived from <bintree> ::= <symbol> has
zero parens.

Program mirrors BNF description mirrors proof:
(+ (count-bt-parens (car bt))
(count-bt-parens (cadr bt))
2))))
<bintree> ::= (<bintree> <bintree>)
Inductive step: Assume that the relation holds for two bin
trees b1, with 2m ≥ 0 parens, and b2, with 2n ≥ 0 parens....
Follow the Grammar!
When defining a program based on structural induction,
the structure of the program should be patterned after the
structure of the data.
Another Point
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Program assumes we're passing it a binary tree: dynamic
type-checking (vs. Java, static / compile-time type check)
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Easier to make type errors:
>
(count-bt-parens '(dont taze me bro))
car: expects argument of type <pair>;
given (dont taze me bro)
Another Example
; remove first occurrence of symbol from
; list of symbols
(define remove-first
(lambda(s los)
(if (null? los)
'()
(if (eqv? (car los) s)
(cdr los)
(cons (car los)
(remove-first
s
(cdr los)))))))
<list-of-symbols> ::= ( ) | (<symbol> . <list-of-symbols> )